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CHAPTER 4 FOURIER SERIES AND INTEGRALS
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318 Chapter 4 Fourier Series and Integrals,Zero comes quickly if we integrate cos mx dx m 0. 0 0 So we use this, Product of sines sin nx sin kx cos n k x cos n k x 4. Integrating cos mx with m n k and m n k proves orthogonality of the sines. The exception is when n k Then we are integrating sin kx 2 12 12 cos 2kx. sin kx sin kx dx dx cos 2kx dx 5,0 0 2 0 2 2, The highlighted term in equation 2 is bk 2 Multiply both sides of 2 by 2. Sine coe cients 2 1,bk S x sin kx dx S x sin kx dx 6. Notice that S x sin kx is even equal integrals from to 0 and from 0 to. I will go immediately to the most important example of a Fourier sine series S x. is an odd square wave with SW x 1 for 0 x It is drawn in Figure 4 1 as. an odd function with period 2 that vanishes at x 0 and x. Figure 4 1 The odd square wave with SW x 2 SW x 1 or 0 or 1. Example 1 Find the Fourier sine coe cients bk of the square wave SW x. Solution For k 1 2 use the rst formula 6 with S x 1 between 0 and. 2 2 cos kx 2 2 0 2 0 2 0,bk sin kx dx 7,0 k 0 1 2 3 4 5 6.
The even numbered coe cients b2k are all zero because cos 2k cos 0 1 The. odd numbered coe cients bk 4 k decrease at the rate 1 k We will see that same. 1 k decay rate for all functions formed from smooth pieces and jumps. Put those coe cients 4 k and zero into the Fourier sine series for SW x. 4 sin x sin 3x sin 5x sin 7x,Square wave SW x 8, Figure 4 2 graphs this sum after one term then two terms and then ve terms You. can see the all important Gibbs phenomenon appearing as these partial sums. 4 1 Fourier Series for Periodic Functions 319, include more terms Away from the jumps we safely approach SW x 1 or 1. At x 2 the series gives a beautiful alternating formula for the number. 4 1 1 1 1 1 1 1 1,1 so that 4 9,1 3 5 7 1 3 5 7, The Gibbs phenomenon is the overshoot that moves closer and closer to the jumps. Its height approaches 1 18 and it does not decrease with more terms of the series. Overshoot is the one greatest obstacle to calculation of all discontinuous functions. like shock waves in uid ow We try hard to avoid Gibbs but sometimes we can t. 4 sin x sin 3x 4 sin x sin 9x,Solid curve 5 terms,4 sin x overshoot. Figure 4 2 Gibbs phenomenon Partial sums 1 bn sin nx overshoot near jumps. Fourier Coe cients are Best, Let me look again at the rst term b1 sin x 4 sin x This is the closest possible.
approximation to the square wave SW by any multiple of sin x closest in the least. squares sense To see this optimal property of the Fourier coe cients minimize the. error over all b1, The error is SW b1 sin x dx The b1 derivative is 2 SW b1 sin x sin x dx. The integral of sin2 x is 2 So the derivative is zero when b1 2 0. S x sin x dx, This is exactly equation 6 for the Fourier coe cient. Each bk sin kx is as close as possible to SW We can nd the coe cients bk one. at a time because the sines are orthogonal The square wave has b2 0 because all. other multiples of sin 2x increase the error Term by term we are projecting the. function onto each axis sin kx,Fourier Cosine Series. The cosine series applies to even functions with C x C x. Cosine series C x a0 a1 cos x a2 cos 2x a0 an cos nx 10. 320 Chapter 4 Fourier Series and Integrals, Every cosine has period 2 Figure 4 3 shows two even functions the repeating. ramp RR x and the up down train UD x of delta functions That sawtooth. ramp RR is the integral of the square wave The delta functions in UD give the. derivative of the square wave For sines the integral and derivative are cosines. RR and UD will be valuable examples one smoother than SW one less smooth. First we nd formulas for the cosine coe cients a0 and ak The constant term a0. is the average value of the function C x,a0 Average a0 C x dx C x dx 11.
I just integrated every term in the cosine series 10 from 0 to On the right side. the integral of a0 is a0 divide both sides by All other integrals are zero. cos nx dx 0 0 0 12, In words the constant function 1 is orthogonal to cos nx over the interval 0. The other cosine coe cients ak come from the orthogonality of cosines As with. sines we multiply both sides of 10 by cos kx and integrate from 0 to. C x cos kx dx a0 cos kx dx a1 cos x cos kx dx ak cos kx 2 dx. You know what is coming On the right side only the highlighted term can be. nonzero Problem 4 1 1 proves this by an identity for cos nx cos kx now 4 has a. plus sign The bold nonzero term is ak 2 and we multiply both sides by 2. Cosine coe cients 2 1,ak C x cos kx dx C x cos kx dx 13. Again the integral over a full period from to also 0 to 2 is just doubled. RR x x 2 x 6 2 x 2 6,Up down U D x,Repeating Ramp RR x. Integral of Square Wave 2 x 2 x, Figure 4 3 The repeating ramp RR and the up down UD periodic spikes are even. The derivative of RR is the odd square wave SW The derivative of SW is U D. 4 1 Fourier Series for Periodic Functions 321, Example 2 Find the cosine coe cients of the ramp RR x and the up down UD x.
Solution The simplest way is to start with the sine series for the square wave. 4 sin x sin 3x sin 5x sin 7x, Take the derivative of every term to produce cosines in the up down delta function. Up down series UD x cos x cos 3x cos 5x cos 7x 14, Those coe cients don t decay at all The terms in the series don t approach zero so. o cially the series cannot converge Nevertheless it is somehow correct and important. Uno cially this sum of cosines has all 1 s at x 0 and all 1 s at x Then. and are consistent with 2 x and 2 x The true way to recognize x is. by the test x f x dx f 0 and Example 3 will do this. For the repeating ramp we integrate the square wave series for SW x and add the. average ramp height a0 2 halfway from 0 to,cos x cos 3x cos 5x cos 7x. Ramp series RR x 15,2 4 12 32 52 72, The constant of integration is a0 Those coe cients. ak drop o like 1 k 2 They could be, computed directly from formula 13 using x cos kx dx but this requires an integration.
by parts or a table of integrals or an appeal to Mathematica or Maple It was much. easier to integrate every sine separately in SW x which makes clear the crucial point. Each degree of smoothness in the function is re ected in a faster decay rate of its. Fourier coe cients ak and bk,No decay Delta functions with spikes. 1 k decay Step functions with jumps,1 k2 decay Ramp functions with corners. 1 k4 decay Spline functions jumps in f, r k decay with r 1 Analytic functions like 1 2 cos x. Each integration divides the kth coe cient by k So the decay rate has an extra. 1 k The Riemann Lebesgue lemma says that ak and bk approach zero for any. continuous function in fact whenever f x dx is nite Analytic functions achieve. a new level of smoothness they can be di erentiated forever Their Fourier series. and Taylor series in Chapter 5 converge exponentially fast. The poles of 1 2 cos x will be complex solutions of cos x 2 Its Fourier series. converges quickly because r k decays faster than any power 1 k p Analytic functions. are ideal for computations the Gibbs phenomenon will never appear. Now we go back to x for what could be the most important example of all. 322 Chapter 4 Fourier Series and Integrals, Example 3 Find the cosine coe cients of the delta function x made 2 periodic. Solution The spike occurs at the start of the interval 0 so safer to integrate from. to We nd a0 1 2 and the other ak 1 cosines because x is even. Average a0 x dx Cosines ak x cos kx dx, Then the series for the delta function has all cosines in equal amounts.
Delta function x cos x cos 2x cos 3x 16, Again this series cannot truly converge its terms don t approach zero But we can graph. the sum after cos 5x and after cos 10x Figure 4 4 shows how these partial sums are. doing their best to approach x They oscillate faster and faster away from x 0. Actually there is a neat formula for the partial sum N x that stops at cos Nx Start. by writing each term 2 cos as ei e i,N 1 2 cos x 2 cos Nx 1 eix e ix eiN x e iN x. This is a geometric progression that starts from e iN x and ends at eiN x We have powers. of the same factor eix The sum of a geometric series is known. Partial sum 1 ei N 2 x e i N 2 x 1 sin N 12 x,up to cos N x 2 eix 2 e ix 2 2 sin 12 x. This is the function graphed in Figure 4 4 We claim that for any N the area underneath. N x is 1 Each cosine integrated from to gives zero The integral of 1 2 is. 1 The central lobe in the graph ends when sin N 12 x comes down to zero and. that happens when N 12 x I think the area under that lobe marked by bullets. approaches the same number 1 18 that appears in the Gibbs phenomenon. In what way does N x approach x The terms cos nx in the series jump around. at each point x,0 not approaching zero At x we see 2 1. 1 2 2 2 and, the sum is 1 2 or 1 2 The bumps in the partial sums don t get smaller than 1 2.
The right test for the delta function x is to multiply by a smooth. f x ak cos kx, and integrate because we only know x from its integrals x f x dx f 0. Weak convergence,N x f x dx a0 aN f 0 18,of N x to x. In this integrated sense weak sense the sums N x do approach the delta function. The convergence of a0 aN is the statement that at x 0 the Fourier series of a. smooth f x ak cos kx converges to the number f 0,4 1 Fourier Series for Periodic Functions 323. 10 x height 21 2,5 x height 11 2,height 1 2,height 1 2. Figure 4 4 The sums N x 1 2 cos x 2 cos Nx 2 try to approach x. Complete Series Sines and Cosines, Over the half period 0 the sines are not orthogonal to all the cosines In fact the.
integral of sin x times 1 is not zero So for functions F x that are not odd or even. we move to the complete series sines plus cosines on the full interval Since our. functions are periodic that full interval can be or 0 2. Complete Fourier series F x a0 an cos nx bn sin nx 19. On every 2 interval all sines and cosines are mutually orthogonal We nd the. Fourier coe cients ak and bk in the usual way Multiply 19 by 1 and cos kx and. sin kx and integrate both sides from to,a0 F x dx ak F x cos kx dx bk F x sin x dx 20. Orthogonality kills o in nitely many integrals and leaves only the one we want. Another approach is to split F x C x S x into an even part and an odd. part Then we can use the earlier cosine and sine formulas The two parts are. F x F x F x F x,C x Feven x S x Fodd x 21, The even part gives the a s and the odd part gives the b s Test on a short square. pulse from x 0 to x h this one sided function is not odd or even. 324 Chapter 4 Fourier Series and Integrals,1 for 0 x h. Example 4 Find the a s and b s if F x square pulse. 0 for h x 2, Solution The integrals for a0 and ak and bk stop at x h where F x drops to zero. The coe cients decay like 1 k because of the jump at x 0 and the drop at x h. Coe cients of square pulse a0 1 dx average,1 h sin kh 1 h 1 cos kh.
ak cos kx dx bk sin kx dx 22, If we divide F x by h its graph is a tall thin rectangle height h1 base h and area 1. When h approaches zero F x h is squeezed into a very thin interval The tall. rectangle approaches weakly the delta function x The average height is area 2. 1 2 Its other coe cients ak h and bk h approach 1 and 0 already known for x. F x ak 1 sin kh 1 bk 1 cos kh,x and 0 as h 0 23,h h kh h kh. When the function has a jump its Fourier series picks the halfway point This. example would converge to F 0 12 and F h 12 halfway up and halfway down. The Fourier series converges to F x at each point where the function is smooth. This is a highly developed theory and Carleson won the 2006 Abel Prize by proving. convergence, for every x except a set of measure zero If the function has nite energy. F x 2 dx he showed that the Fourier series converges almost everywhere. Energy in Function Energy in Coe cients, There is an extremely important equation the energy identity that comes from. integrating F x 2 When we square the Fourier series of F x and integrate from. to all the cross terms drop out The only nonzero integrals come from 12. and cos2 kx and sin2 kx multiplied by a20 and a2k and b2k. Energy in F x a0 ak cos kx bk sin kx 2 dx,F x 2dx 2 a20 a21 b21 a22 b22.
The energy in F x equals the energy in the coe cients The left side is like the. length squared of a vector except the vector is a function The right side comes from. an in nitely long vector of a s and b s The lengths are equal which says that the. Fourier transform, from function to vector is like an orthogonal matrix Normalized. by constants 2 and we have an orthonormal basis in function space. What is this function space It is like ordinary 3 dimensional space except the. are functions Their length f comes from integrating instead of adding. f 2 f x 2 dx These functions ll Hilbert space The rules of geometry hold. 4 1 Fourier Series for Periodic Functions 325, Length f 2 f f comes from the inner product f g f x g x dx. Orthogonal functions f g 0 produce a right triangle f g 2 f 2 g 2. I have tried to draw Hilbert space in Figure 4 5 It has in nitely many axes The. energy identity 24 is exactly the Pythagoras Law in in nite dimensional space. cos kx sin x,v2k f A0 v0 A1 v1 B1 v2,function in Hilbert space. f 2 A20 A21 B12,1 vi vj 0 cos x, Figure 4 5 The Fourier series is a combination of orthonormal v s sines and cosines. Complex Exponentials ck eikx, This is a small step and we have to take it In place of separate formulas for a0 and ak.
and bk we will have one formula for all the complex coe cients ck And the function. F x might be complex as in quantum mechanics The Discrete Fourier Transform. will be much simpler when we use N complex exponentials for a vector We practice. in advance with the complex in nite series for a 2 periodic function. Complex Fourier series F x c0 c1 eix c 1 e ix cn einx 25. If every cn c n we can combine einx with e inx into 2 cos nx Then 25 is the. cosine series for an even function If every cn c n we use einx e inx 2i sin nx. Then 25 is the sine series for an odd function and the c s are pure imaginary. To nd ck multiply 25 by e ikx not eikx and integrate from to. F x e ikx dx c0 e ikx dx c1 eix e ikx dx ckeikxe ikxdx. The complex exponentials are orthogonal Every integral on the right side is zero. except for the highlighted term when n k and eikx e ikx 1 The integral of 1 is. 2 That surviving term gives the formula for ck,Fourier coe cients F x e ikx dx 2 ck for k 0 1 26. 326 Chapter 4 Fourier Series and Integrals, Notice that c0 a0 is still the average of F x because e0 1 The orthogonality. of einx and eikx is checked by integrating as always But the complex inner product. F G takes the complex conjugate G of G Before integrating change eikx to e ikx. Complex inner product Orthogonality of einx and eikx. i n k x e 27,F G F x G x dx e dx 0, Example 5 Add the complex series for 1 2 eix and 1 2 e ix These geometric. series have exponentially fast decay from 1 2k The functions are analytic. 1 eix e2ix 1 e ix e 2ix cos x cos 2x cos 3x,2 4 8 2 4 8 2 4 8. When we add those functions we get a real analytic function. 1 1 2 e ix 2 eix 4 2 cos x,2 eix 2 e ix 2 e 2 e,ix ix 5 4 cos x.
This ratio is the in nitely smooth function whose cosine coe cients are 1 2 k. 1 for s x s h, Example 6 Find ck for the 2 periodic shifted pulse F x. 0 elsewhere in, Solution The integrals 26 from to become integrals from s to s h. 1 ikx 1 e ikx iks 1 e ikh,ck 1 e dx e 29,2 s 2 ik s 2 ik. Notice above all the simple e ect of the shift by s It modulates each c k by e iks The. energy is unchanged the integral of F 2 just shifts and all e iks 1. Shift F x to F x s Multiply ck by e iks 30, Example 7 Centered pulse with shift s h 2 The square pulse becomes centered. around x 0 This even function equals 1 on the interval from h 2 to h 2. 1 e ikh 1 sin kh 2,Centered by s h2 ck eikh 2,2 ik 2 k 2.
Divide by h for a tall pulse The ratio of sin kh 2 to kh 2 is the sinc function. Fcentered kh ikx 1 h for h 2 x h 2,Tall pulse sinc e. h 2 2 0 elsewhere in, That division by h produces area 1 Every coe cient approaches 2 1. The Fourier series for the tall thin pulse again approaches the Fourier series for x. 4 1 Fourier Series for Periodic Functions 327, Hilbert space can contain vectors c c0 c1 c 1 c2 c 2 instead of functions. F x The length of c is 2 ck 2 F 2dx The function space is often denoted. by L2 and the vector space is 2 The energy identity is trivial but deep Integrating. the Fourier series for F x times F x orthogonality kills every cn ck for n. leaves the ck ck ck 2,F x 2dx cn einx ck e ikx dx 2 c0 2 c1 2 c 1 2 31. This is Plancherel s identity The energy in x space equals the energy in k space. Finally I want to emphasize the three big rules for operating on F x ck e. 1 The derivative has Fourier coe cients ikck energy moves to high k. 2 The integral of F x has Fourier coe cients k,0 faster decay.
3 The shift to F x s has Fourier coe cients e iks ck no change in energy. Application Laplace s Equation in a Circle, Our rst application is to Laplace s equation The idea is to construct u x y as an. in nite series choosing its coe cients to match u0 x y along the boundary Every. thing depends on the shape of the boundary and we take a circle of radius 1. Begin with the simple solutions 1 r cos r sin r 2 cos 2 r 2 sin 2 to Laplace s. equation Combinations of these special solutions give all solutions in the circle. u r a0 a1 r cos b1 r sin a2 r 2 cos 2 b2 r 2 sin 2 32. It remains to choose the constants ak and bk to make u u0 on the boundary. For a circle u0 is periodic since and 2 give the same point. Set r 1 u0 a0 a1 cos b1 sin a2 cos 2 b2 sin 2 33, This is exactly the Fourier series for u0 The constants ak and bk must be the. Fourier coe cients of u0 Thus the problem is completely solved if an in nite. series 32 is acceptable as the solution, Example 8 Point source u0 at 0 The whole boundary is held at u0 0. except for the source at x 1 y 0 Find the temperature u r inside. Fourier series for u0 cos cos 2 cos 3 e,328 Chapter 4 Fourier Series and Integrals. Inside the circle each cos n is multiplied by r n, In nite series for u u r r cos r 2 cos 2 r 3 cos 3 34.
Poisson managed to sum this in nite series It involves a series of powers of re i. So we know the response at every r to the point source at r 1 0. Temperature inside circle u r 35,2 1 r 2 2r cos, At the center r 0 this produces the average of u0 which is a0 1 2 On the. boundary r 1 this produces u 0 except at the point source where cos 0 1. 1 1 r2 1 1 r,On the ray 0 u r 36,2 1 r 2 2r 2 1 r, As r approaches 1 the solution becomes in nite as the point source requires. Example 9 Solve for any boundary values u0 by integrating over point sources. When the point source swings around to angle the solution 35 changes from to. Integrate this Green s function to solve in the circle. Poisson s formula u r u0 d 37,2 1 r 2 2r cos, Ar r 0 the fraction disappears and u is the average u0 d 2 The steady. state temperature at the center is the average temperature around the circle. Poisson s formula illustrates a key idea Think of any u 0 as a circle of point sources. The source at angle produces the solution inside the integral 37 Integrating. around the circle adds up the responses to all sources and gives the response to u 0. Example 10 u0 1 on the top half of the circle and u0 1 on the bottom half. Solution The boundary values are the square wave SW Its sine series is in 8. 4 sin sin 3 sin 5,Square wave for u0 SW 38, Inside the circle multiplying by r r 2 r 3 gives fast decay of high frequencies. 4 r sin r 3 sin 3 r 5 sin 5,Rapid decay inside u r 39.
Laplace s equation has smooth solutions even when u0 is not smooth. 4 1 Fourier Series for Periodic Functions 329,WORKED EXAMPLE. A hot metal bar is moved into a freezer zero temperature The sides of the bar. are coated so that heat only escapes at the ends What is the temperature u x t. along the bar at time t It will approach u 0 as all the heat leaves the bar. Solution The heat equation is ut uxx At t 0 the whole bar is at a constant. temperature say u 1 The ends of the bar are at zero temperature for all time t 0. This is an initial boundary value problem, Heat equation ut uxx with u x 0 1 and u 0 t u t 0 40. Those zero boundary conditions suggest a sine series Its coe cients depend on t. Series solution of the heat equation u x t bn t sin nx 41. The form of the solution shows separation of variables In a comment below we. look for products A x B t that solve the heat equation and the boundary conditions. What we reach is exactly A x sin nx and the series solution 41. Two steps remain First choose each bn t sin nx to satisfy the heat equation. bn t sin nx n2 bn t sin nx,Substitute into ut uxx bn t e n t bn 0. Notice bn n2 bn Now determine each bn 0 from the initial condition u x 0 1. on 0 Those numbers are the Fourier sine coe cients of SW x in equation 38. Box function square wave bn 0 sin nx 1 bn 0 for odd n. This completes the series solution of the initial boundary value problem. Bar temperature u x t e n t sin nx 42, For large n high frequencies the decay of e n t is very fast The dominant term. 4 e t sin x for large times will come from n 1 This is typical of the heat. equation and all di usion that the solution the temperature pro le becomes very. smooth as t increases, Numerical di culty I regret any bad news in such a beautiful solution To compute.
u x t we would probably truncate the series in 42 to N terms When that nite. series is graphed on the website serious bumps appear in uN x t You ask if there. is a physical reason but there isn t The solution should have maximum temperature. at the midpoint x 2 and decay smoothly to zero at the ends of the bar. 330 Chapter 4 Fourier Series and Integrals, Those unphysical bumps are precisely the Gibbs phenomenon The initial. u x 0 is 1 on 0 but its odd re ection is 1 on 0 That jump has produced. the slow 4 n decay of the coe cients with Gibbs oscillations near x 0 and x. The sine series for u x t is not a success numerically Would nite di erences help. Separation of variables We found bn t as the coe cient of an eigenfunction sin nx. Another good approach is to put u A x B t directly into ut uxx. Separation A x B t A x B t requires constant 43, A A is constant in space B B is constant in time and they are equal. gives A sin x and cos x gives B e t, The products AB e t sin x and e t cos x solve the heat equation for any. number But the boundary condition u 0 t 0 eliminates the cosines Then. u t 0 requires n2 1 4 9 to have sin 0 Separation of variables. has recovered the functions in the series solution 42. Finally u x 0 1 determines the numbers 4 n for odd n We nd zero for even. n because sin nx has n 2 positive loops and n 2 negative loops For odd n the extra. positive loop is a fraction 1 n of all loops giving slow decay of the coe cients. Heat bath the opposite problem The solution on the website is 1 u x t. because it solves a di erent problem The bar is initially frozen at U x 0. 0 It is placed into a heat bath at the xed temperature U 1 or U T0. The new unknown is U and its boundary conditions are no longer zero. The heat equation and its boundary conditions are solved rst by UB x t In. this example UB 1 is constant Then the di erence V U UB has zero boundary. values and its initial values are V 1 Now the eigenfunction method or sepa. ration of variables solves for V The series in 42 is multiplied by 1 to account. for V x 0 1 Adding back UB solves the heat bath problem U UB V. Here UB 1 is the steady state solution at t and V is the transient solution. The transient starts at V 1 and decays quickly to V 0. Heat bath at one end The website problem is di erent in another way too The. Dirichlet condition u t 1 is replaced by the Neumann condition u 1 t 0. Only the left end is in the heat bath Heat ows down the metal bar and out at the. far end now located at x 1 How does the solution change for xed free. Again UB 1 is a steady state The boundary conditions apply to V 1 UB. Fixed free 1,V 0 0 and V 1 0 lead to A x sin n x 44. eigenfunctions 2,4 1 Fourier Series for Periodic Functions 331.
Those eigenfunctions give a new form for the sum of Bn t An x. n 21 2 2 t 1,Fixed free solution V x t Bn 0 e sin n x 45. All frequencies shift by 12 and multiply by because A A has a free end. at x 1 The crucial,is Does orthogonality still hold for these new. eigenfunctions sin n 12 x on 0 1 The answer is yes because this xed free. Sturm Liouville problem A A is still symmetric, Summary The series solutions all succeed but the truncated series all fail We can. see the overall behavior of u x t and V x t But their exact values close to the. jumps are not computed well until we improve on Gibbs. We could have solved the xed free problem on 0 1 with the xed xed solution. on 0 2 That solution will be symmetric around x 1 so its slope there is zero. Then rescaling x by 2 changes sin n 12 x into sin 2n 1 x I hope you like the. graphics created by Aslan Kasimov on the cse website. Problem Set 4 1,1 Find the Fourier series on x for. a f x sin3 x an odd function,b f x sin x an even function.
d f x ex using the complex form of the series, What are the even and odd parts of f x ex and f x eix. 2 From Parseval s formula the square wave sine coe cients satisfy. b21 b22 f x 2 dx 1 dx 2,Dirive the remarkable sum 2 8 1 91 1. 3 If a square pulse is centered at x 0 to give,f x 1 for x f x 0 for x. draw its graph and nd its Fourier coe cients ak and bk. 4 Suppose f has period T instead of 2x so that f x f x T Its graph from. T 2 to T 2 is repeated on each successive interval and its real and complex. Fourier series are,f x a0 a1 cos b1 sin ck eik2 x T. Multiplying by the right functions and integrating from T 2 to T 2 nd ak.

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